Question

在Java中，我有2个字节：

byte b1 = (byte) 0b11111111, b2 = (byte) 0b00000000;

我想混合它们，以便每个第一位来自b1，而另一位来自b2（从左到右阅读）。输入的前半部分和后半部分是分开完成的，因此结果是2个字节。结果b3和b4如下所示。

byte b3 = (byte) 0b10101010, b4 = 0b10101010;

说明这些位是如何唯一的（使用字母指定唯一位）：

byte b1 = (byte) 0bHGFEDCBA, b2 = (byte) 0bPONMLKJI;

结果将是：

byte b3 = (byte) 0bHPGOFNEM, b4 = 0bDLCKBJAI;

或者，图形上，

    +---+---+---+---+---+---+---+---+
b1  | H | G | F | E | D | C | B | A |
    +---+---+---+---+---+---+---+---+
      |   |   |   |   |   |   |   | 
      |   |   |   |   |   |   |   +--------------------------------------------+
      |   |   |   |   |   |   +----------------------------------------+       |
      |   |   |   |   |   +------------------------------------+       |       |
      |   |   |   |   +--------------------------------+       |       |       |
      |   |   |   +-------------------+                |       |       |       |  
      |   |   +---------------+       |                |       |       |       |  
      |   +-----------+       |       |                |       |       |       | 
      +-------+       |       |       |                |       |       |       | 
              |       |       |       |                |       |       |       |  
            +---+---+---+---+---+---+---+---+        +---+---+---+---+---+---+---+---+
         b3 | H | P | G | O | F | N | E | M |     b4 | D | L | C | K | B | J | A | I |
            +---+---+---+---+---+---+---+---+        +---+---+---+---+---+---+---+---+
                  |       |       |       |                |       |       |       |
      +-----------+       |       |       |                |       |       |       |
      |   +---------------+       |       |                |       |       |       |
      |   |   +-------------------+       |                |       |       |       |
      |   |   |   +-----------------------+                |       |       |       |
      |   |   |   |   +------------------------------------+       |       |       |
      |   |   |   |   |   +----------------------------------------+       |       |
      |   |   |   |   |   |   +--------------------------------------------+       |
      |   |   |   |   |   |   |   +------------------------------------------------+
      |   |   |   |   |   |   |   | 
    +---+---+---+---+---+---+---+---+
b2  | P | O | N | M | L | K | J | I |
    +---+---+---+---+---+---+---+---+

实现这一目标的最简单方法是什么？

Answer 1

正如你所说，如果你的心脏被放在一个单线上，那该怎么样：

public static int interleave(short b1, short b2) {
    return((int)(((b2 * 0x0101010101010101L & 0x8040201008040201L) * 
        0x0102040810204081L >> 49) & 0x5555) |
        (int)(((b1 * 0x0101010101010101L & 0x8040201008040201L) * 
        0x0102040810204081L >> 48) & 0xAAAA));
}

这将返回带有b3＆amp;的int。 b4作为低16位，你可以移位和掩盖：

int b3b4 = interleave(b1, b2);
int b3 = b3b4 >> 8;
int b4 = b3b4 & 0b11111111;

算法由Interleave bits with 64-bit multiply

提供

Answer 2

首先，创建一个在一个字节中扩展位的方法，并返回一个int，其中最后16位设置为原始字节的位：

static int spread(int b) {
    int res = 0;
    for (int i = 0 ; i != 8 ; i++) {
        if ((b & 1<<i) != 0) {
            res |= 1<<(2*i);
        }
    }
    return res;
}

使用这个方法，通过对第一个点差的结果进行OR运算产生结果，第二个点差的结果向左移一个：

int res = spread(b1) | (spread(b2) << 1);

由于您的号码很小，您可以为所有256种可能性预先计算spread(x)。

这produces Morton's table。将其复制到您的班级，并使您的解决方案成为一个单行：

int res = morton[b1] | (morton[b2] << 1);

// This declaration goes at the bottom of your file.
// The numbers are copied from the program output at the link above:
private static final short[] morton = new short[] {
    0x0000, 0x0001, 0x0004, 0x0005, 0x0010, 0x0011, 0x0014, 0x0015, 
    0x0040, 0x0041, 0x0044, 0x0045, 0x0050, 0x0051, 0x0054, 0x0055, 
    0x0100, 0x0101, 0x0104, 0x0105, 0x0110, 0x0111, 0x0114, 0x0115, 
    0x0140, 0x0141, 0x0144, 0x0145, 0x0150, 0x0151, 0x0154, 0x0155, 
    0x0400, 0x0401, 0x0404, 0x0405, 0x0410, 0x0411, 0x0414, 0x0415, 
    0x0440, 0x0441, 0x0444, 0x0445, 0x0450, 0x0451, 0x0454, 0x0455, 
    0x0500, 0x0501, 0x0504, 0x0505, 0x0510, 0x0511, 0x0514, 0x0515, 
    0x0540, 0x0541, 0x0544, 0x0545, 0x0550, 0x0551, 0x0554, 0x0555, 
    0x1000, 0x1001, 0x1004, 0x1005, 0x1010, 0x1011, 0x1014, 0x1015, 
    0x1040, 0x1041, 0x1044, 0x1045, 0x1050, 0x1051, 0x1054, 0x1055, 
    0x1100, 0x1101, 0x1104, 0x1105, 0x1110, 0x1111, 0x1114, 0x1115, 
    0x1140, 0x1141, 0x1144, 0x1145, 0x1150, 0x1151, 0x1154, 0x1155, 
    0x1400, 0x1401, 0x1404, 0x1405, 0x1410, 0x1411, 0x1414, 0x1415, 
    0x1440, 0x1441, 0x1444, 0x1445, 0x1450, 0x1451, 0x1454, 0x1455, 
    0x1500, 0x1501, 0x1504, 0x1505, 0x1510, 0x1511, 0x1514, 0x1515, 
    0x1540, 0x1541, 0x1544, 0x1545, 0x1550, 0x1551, 0x1554, 0x1555, 
    0x4000, 0x4001, 0x4004, 0x4005, 0x4010, 0x4011, 0x4014, 0x4015, 
    0x4040, 0x4041, 0x4044, 0x4045, 0x4050, 0x4051, 0x4054, 0x4055, 
    0x4100, 0x4101, 0x4104, 0x4105, 0x4110, 0x4111, 0x4114, 0x4115, 
    0x4140, 0x4141, 0x4144, 0x4145, 0x4150, 0x4151, 0x4154, 0x4155, 
    0x4400, 0x4401, 0x4404, 0x4405, 0x4410, 0x4411, 0x4414, 0x4415, 
    0x4440, 0x4441, 0x4444, 0x4445, 0x4450, 0x4451, 0x4454, 0x4455, 
    0x4500, 0x4501, 0x4504, 0x4505, 0x4510, 0x4511, 0x4514, 0x4515, 
    0x4540, 0x4541, 0x4544, 0x4545, 0x4550, 0x4551, 0x4554, 0x4555, 
    0x5000, 0x5001, 0x5004, 0x5005, 0x5010, 0x5011, 0x5014, 0x5015, 
    0x5040, 0x5041, 0x5044, 0x5045, 0x5050, 0x5051, 0x5054, 0x5055, 
    0x5100, 0x5101, 0x5104, 0x5105, 0x5110, 0x5111, 0x5114, 0x5115, 
    0x5140, 0x5141, 0x5144, 0x5145, 0x5150, 0x5151, 0x5154, 0x5155, 
    0x5400, 0x5401, 0x5404, 0x5405, 0x5410, 0x5411, 0x5414, 0x5415, 
    0x5440, 0x5441, 0x5444, 0x5445, 0x5450, 0x5451, 0x5454, 0x5455, 
    0x5500, 0x5501, 0x5504, 0x5505, 0x5510, 0x5511, 0x5514, 0x5515, 
    0x5540, 0x5541, 0x5544, 0x5545, 0x5550, 0x5551, 0x5554, 0x5555
};

如何在一个字节中扩展位？

2 个答案: